integral from 0 to pi/2 of arctan(2x)

Lösung

\int_{0}^{\frac{π}{2}} arctan (2 x) d x

\frac{1}{4} (2 π arctan (π) - ln (π^{2} + 1))

Dezimale

1.38683 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} arctan (2 x) d x

Wende die partielle Integration an

= [x arctan (2 x) - \int \frac{2 x}{4 x ^{2} + 1} d x]_{0}^{\frac{π}{2}}

\int \frac{2 x}{4 x ^{2} + 1} d x = \frac{1}{4} ln 4 x^{2} + 1

= [x arctan (2 x) - \frac{1}{4} ln 4 x^{2} + 1]_{0}^{\frac{π}{2}}

Vereinfache

[x arctan (2 x) - \frac{1}{4} ln 4 x^{2} + 1]_{0}^{\frac{π}{2}} : [\frac{1}{4} (4 x arctan (2 x) - ln 4 x^{2} + 1)]_{0}^{\frac{π}{2}}

= [\frac{1}{4} (4 x arctan (2 x) - ln 4 x^{2} + 1)]_{0}^{\frac{π}{2}}

Berechne die Grenzen:

\frac{1}{4} (2 π arctan (π) - ln (π^{2} + 1))

= \frac{1}{4} (2 π arctan (π) - ln (π^{2} + 1))

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